A bottler of drinking water fills plastic bottles with a mean volume of 1,006 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. What proportion of bottles have
Volumes greater than 1,009 mL?
A) 1.0000
B) 0.9998
C) 0.6331
D) 0.3085

Question

Quizplus · Accepted Answer

To find the proportion of bottles with volumes greater than 1,009 mL, we use the Z-score formula: Z = (X - μ) / σ, where X is the value in question (1,009 mL), μ is the mean (1,006 mL), and σ is the standard deviation (6 mL). Plugging in the values gives Z = (1009 - 1006) / 6 = 0.5. Looking up 0.5 on the standard normal distribution table gives a left-tail value of approximately 0.6915, meaning the right tail (greater than 1,009 mL) is 1 - 0.6915 = 0.3085.

A Bottler of Drinking Water Fills Plastic Bottles with a Mean